I have to solve given task: There is football tournament with 8 groups by 4 teams where everyone plays vs everyone 2 times. In round two only 2 teams are qualified and they play again two matches. In all next rounds two matches are played till there is only one winner. How many matches are played?
My logic:
round 1 $C\frac{3}{4} = 12.8 = 96$
round 2 $C\frac{2}{2} = 4.8 = 32$
round 3 $C\frac{1}{2} = 2.8 = 16$
round 4 $C\frac{2}{2} = 1.8 = 8$
round 4 $C\frac{2}{2} = 1.4 = 4$
round 4 $C\frac{2}{2} = 1.2 = 2$
$96 + 32 + 16 + 8 + 4 + 2 + 1 = 159$
Are my calculations correct? Is there any formula easy way to calculate it?
I will use the binomial coefficient notation $$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$ for the number of ways of selecting a subset of $k$ elements from a set with $n$ elements. It corresponds to your notation $C\frac{k}{n}$.
Round-robin qualification round: Since each match involves two teams, the number of ways for each team in a four-team group to play each other once is $$\binom{4}{2}$$ Since each team in the four-team groups plays each other team twice, the number of matches played in each group is $$2\binom{4}{2}$$ Since there are eight groups of four teams, the number of matches played in the round-robin qualification round is $$8 \cdot 2\binom{4}{2} = 8 \cdot 2 \cdot 6 = 96$$ as you found.
First elimination round: Since two teams qualify from each of the eight groups, there are $8 \cdot 2 = 16$ teams in this round. Each team is matched with one other team, forming eight pairs of opponents. In each of those eight pairs, the teams play each other twice, giving $8 \cdot 2 = 16$ matches in this round.
Quarter finals: (second elimination round) Since the eight winners of the first elimination round are left, they form four pairs of opponents. In each pair, the teams play each other twice. Thus, there are $4 \cdot 2 = 8$ matches in this round.
Semi-finals: (third elimination round) Since the four winners of the quarter finals are left, they form two pairs of opponents. In each pair, the teams play each other twice. Thus, there are $2 \cdot 2 = 4$ matches in this round.
Finals: (fourth elimination round) Since the two winners of the semi-finals are left, they play each other twice for the championship. There are $2$ matches in this round.
Total: The total number of matches played in the tournament is $$96 + 16 + 8 + 4 + 2 = 126$$
In counting the elimination matches, you counted each match twice. One possible reason for this is that you may not have taken into account that each match involves two teams.